Tìm x a.x^3-9x =0 b.x^2+4x+4-y^2 c. x^2-2xy+7x-14y
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a) \(xy^2-2xy+x=x\left(y^2-2y+1\right)=x\left(y-1\right)^2\)
b) \(x^2-xy+x-y=x\left(x-y\right)+\left(x-y\right)=\left(x+1\right)\left(x-y\right)\)
a) \(x^4-2x^2=0\)
⇔ \(x^2\left(x^2-2\right)=0\)
⇔ \(x^2\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)=0\)
⇔ \(\left[{}\begin{matrix}x=0\\x=\pm\sqrt{2}\end{matrix}\right.\)
b) \(x^2+3x+2=0\)
\(\Leftrightarrow x\left(x+1\right)+2\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-2\end{matrix}\right.\)
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1) \(5x-5y+x\left(x-y\right)\)
\(=5\left(x-y\right)+x\left(x-y\right)\)
\(=\left(x-y\right)\left(x+5\right)\)
2) \(x^2+4x+3\)
\(=\left(x^2+x\right)+\left(3x+3\right)\)
\(=x\left(x+1\right)+3\left(x+1\right)\)
\(=\left(x+1\right)\left(x+3\right)\)
3) \(x^2-2xy+y^2-z^2\)
\(=\left(x-y\right)^2-z^2\)
\(=\left(x-y-z\right)\left(x-y+z\right)\)
4) \(x\left(x-5\right)-3x+15\)
\(=x\left(x-5\right)-3\left(x-5\right)\)
\(=\left(x-5\right)\left(x-3\right)\)
5) \(y^2-x^2+2x-1\)
\(=y^2-\left(x^2-2x+1\right)\)
\(=y^2-\left(x-1\right)^2\)
\(=\left(x+y-1\right)\left(y-x+1\right)\)
\(1.\left(x-y\right)\left(x+5\right)\)
\(2.\left(x+1\right)\left(x+3\right)\)
\(3.\left(x-y-z\right)\left(x-y+z\right)\)
\(4.\left(x-3\right)\left(x-5\right)\)
\(5.\left(y-x+1\right)\left(y+x+1\right)\)
\(7.\left(x+1\right)\left(x-2\right)^2\)
\(8.\left(x-5\right)\left(x+3\right)\)
\(10.\left(y+1\right)\left(2x+z\right)\)
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1)
5x - 5y + x ( x - y ) = (x-y)(5+x)
2)
x2+4x+3=x2+x+3x+3=(x+1)(x+3)
3)x2-2xy+y2-z2=\(\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\)
4)\(x\left(x-5\right)-3x+15=\left(x-3\right)\left(x-5\right)\)
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a/ x^3-6x^2+12x-8
=(x-2)^3
b/x^2+5x+4
=x^2+x+4x+4
=x(x+1)+4(x+1)
=(x+1)(x+4)
c/ 16^2-9(x+1)^2=0
<=> (4x-3x-3)(4x+3x+3)=0
<=>x-3=0 hay 7x+3=0
<=> x=3 hay x=-3/7
d/ x^3-2x^2-x+2
=x^2(x-2)-(x-2)
=(x-2)(x^2-1)
=(x-2)(x-1)(x+1)
e/x^2+y^2-2xy-x+y
=(x-y)^2-(x-y)
f/x^3+y^3+3y^2+3y+1
=x^3+(y+1)^3
=(x+y+1)[x^2-xy-x+(y+1)^2]
=(x+y+1)(x^2-xy-x+y^2+2y+1)
b. x2+2.5/2x+(5/2)2-(5/2)2+4
= (x+5/2)2-25/4+4
=(x+5/2)2-(3/2)2
= x+ 5/2 -3/2 ) . (x+5/2-3/2)
= (x+1 ) (x+2)
c.
(4x)2- [3(x+1)]2 =0
[4x-3(x+1)] [4x+3(x+1)] =0
(x-3) (7x+3) =0
<=> x-3 =0 => x = 3
7x+3=0 => x= -3/7
d. x3-2x2-x+2
= (x3-2x2) - (x+2)
= x2 (x-2) - (x-2)
= (x-2) (x2-1)
CHÚC BẠN HỌC TỐT
* Tớ còn a, e, và f sorry nó k dễ để suy nghĩ trong thơi gian ngắn được nên tớ bỏ !! Ahihihih
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\(a,=x^2-4x+4-\dfrac{15}{4}=\left(x-2\right)^2-\dfrac{15}{4}=\left(x-2-\dfrac{\sqrt{15}}{2}\right)\left(x-2+\dfrac{\sqrt{15}}{2}\right)\\ b,=?\\ c,\Rightarrow x^2+7x-8=0\\ \Rightarrow\left(x+8\right)\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=-8\\x=1\end{matrix}\right.\\ d,Sửa:x^3-3x^2=-27+9x\\ \Rightarrow x^3-3x^2+9x-27=0\\ \Rightarrow x^2\left(x-3\right)+9\left(x-3\right)=0\\ \Rightarrow\left(x^2+9\right)\left(x-3\right)=0\Rightarrow\left[{}\begin{matrix}x^2=-9\left(vô.lí\right)\\x=3\end{matrix}\right.\\ \Rightarrow x=3\\ e,\Rightarrow x\left(x-3\right)-7x+21=0\\ \Rightarrow x\left(x-3\right)-7\left(x-3\right)=0\\ \Rightarrow\left(x-7\right)\left(x-3\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=3\end{matrix}\right.\\ f,\Rightarrow x^2\left(x-2\right)+\left(x-2\right)=0\\ \Rightarrow\left(x^2+1\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=2\end{matrix}\right.\\ \Rightarrow x=2\)
\(g,\Rightarrow x^2-4x+4=0\\ \Rightarrow\left(x-2\right)^2=0\\ \Rightarrow x=2\\ h,Sửa:x^3-x^2+x=1\\ \Rightarrow x^2\left(x-1\right)+\left(x-1\right)=0\\ \Rightarrow\left(x^2+1\right)\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=1\end{matrix}\right.\\ \Rightarrow x=1\)
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a) Ta có :\(x^3-3.x^2.2+3.x+2^2+2^3\)(Hằng đẳng thức số 5 đấy bạn)
=\(\left(x-2\right)^3\)
b) Ta có:\(x^2+5x+4=x^2+4x+1x+4\)
\(=\left(x^2+4x+4\right)+x\)
\(=\left(x+2\right)^2+x\)
c) Ta có :\(16x^2-9\left(x+1\right)^2=0\)
\(\left[4x+3\left(x+1\right)\right].\left[4x-3\left(x+1\right)\right]=0\)(Hằng đẩng thức số 3)
\(\left(4x+3x+3\right).\left(4x-3x-3\right)=0\)
\(7x+3.\left(x-3\right)=0\)
\(\Rightarrow7x+3=0\)hoặc \(x-3=0\)
\(\Rightarrow7x=-3\) hoặc \(x=0+3\)
\(\Rightarrow x=\frac{-3}{7}\) hoặc \(x=3\)
Vậy:\(x=\frac{-3}{7};3\)
d) Ta có \(x^3-2x^2-x+2\)
\(=\left(x^3-x\right)-\left(2x^2-2\right)\)
\(=x\left(x^2-1\right)-2\left(x^2-1\right)\)
\(=\left(x-2\right).\left(x^2-1\right)\)
Bây giờ hơi trễ rồi để mai mình làm tiếp 2 câu cuối nhá.
Rất vui khi được giúp bạn !!1 : =))
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c: \(x^2-4x^2y^2+y^2+2xy\)
\(=\left(x+y\right)^2-\left(2xy\right)^2\)
\(=\left(x+y+2xy\right)\left(x+y-2xy\right)\)
d: \(7x^2-14xy^2+7y^4\)
\(=7\left(x^2-2xy^2+y^4\right)\)
\(=7\left(x-y^2\right)^2\)
a ) x3 - 9x=0
<=> x (x2 - 3 )= 0
<=> x(x+3)(x-3)
<=> x=0
hoặc x=0-3=-3
hoặc x=0+3=3